package algorithm.problems.backtracking;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * Created by gouthamvidyapradhan on 14/03/2017.
 * Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
 * <p>
 * Each number in C may only be used once in the combination.
 * <p>
 * Note:
 * All numbers (including target) will be positive integers.
 * The solution set must not contain duplicate combinations.
 * For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
 * A solution set is:
 * [
 * [1, 7],
 * [1, 2, 5],
 * [2, 6],
 * [1, 1, 6]
 * ]
 */

public class CombinationSumII {
    /**
     * Main method
     *
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception {
        int[] candidates = {1, 1, 2, 2};
        List<List<Integer>> result = new CombinationSumII().combinationSum2(candidates, 4);
    }

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<>();
        combination(0, target, candidates, new ArrayList<>(), result);
        return result;
    }

    private void combination(int i, int target, int[] candidates, List<Integer> row, List<List<Integer>> result) {
        if (target == 0) {
            result.add(new ArrayList<>(row));
        } else if (target > 0) {
            for (int j = i, l = candidates.length; j < l; j++) {
                if (j > i && candidates[j] == candidates[j - 1]) continue;
                row.add(candidates[j]);
                combination(j + 1, target - candidates[j], candidates, row, result);
                row.remove(row.size() - 1);
            }
        }
    }
}
